Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0 $ and $k \geq 0$: $${n \choose k} = (-1)^k {-n + k - 1 \choose k}$$

So now let's look at one case for using the binomial coefficient: $$(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$$

How do I evaluate $\sum_{k = 0}^{n}$ when $n < 0$? From searching around on the internet I think it's just an infinite series, i.e. $k$ keeps incrementing by 1 forever. But that gets me confused about

$$\begin{align*} (a + b)^n &= a^n(1 + \frac{b}{a})^n \\ &= a^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{b}{a}\right)^k\right)\\ &= a^n \left(1 + n \left(\frac{b}{a}\right) + \frac{(n)(n-1)}{2}\left(\frac{b}{a}\right)^2 + \cdots\right) \end{align*}$$ and $$\begin{align*} (b + a)^n &= b^n\left(1 + \frac{a}{b}\right)^n\\ &= b^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{a}{b}\right)^k\right)\\ &= b^n \left(1 + n \left(\frac{a}{b}\right) + \frac{(n)(n-1)}{2}\left(\frac{a}{b}\right)^2 + \cdots\right) \end{align*}$$

Now the two should be equal, but in the first sum I'd never get a $b^n$ and in the second sum I'd never get a $a^n$?

share|cite|improve this question
The first sum only converges if $\left| \frac{a}{b} \right| > 1$ whereas the second sum only converges if this is less than $1$, so the two series are never simultaneously valid. – Qiaochu Yuan Nov 26 '11 at 4:36
Ok, so my interpretation of $\sum_{k = 0}^{n}$ is correct for $n < 0$? I see that you are correct. Does this mean that the formula $(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$ is valid for all x is $n \geq 0$ and only for $|x| < 1$ when $n < 0$? I have not seen this condition before for this formula. – Tianxiang Xiong Nov 26 '11 at 4:43
Correct. In fact $n$ may be any complex number and the sum will be infinite unless $n$ is a non-negative integer. – Qiaochu Yuan Nov 26 '11 at 4:47
Ok. I suppose that this should also be extended to $(x+y)^n$ as well, where it equals $\sum_{k=0}^n {n \choose k}x^{n-k}y^k$ iff $|\frac{y}{x}| < 1$ and $\sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$ iff $|\frac{x}{y}| < 1$? Thanks for your help. – Tianxiang Xiong Nov 26 '11 at 4:55
Your sums should go from 0 to $\infty$, not 0 to $n$. – Ted Nov 26 '11 at 5:51
up vote 6 down vote accepted

The below is too long for a comment so I'm including it here even though I'm not sure it "answers" the question.

If you think about $(1+x)^{-n}$ as living in the ring of formal power series $\mathbb{Z}[[x]]$, then you can show that $$(1+x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$ and the identity $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ seems very natural. Here's how...

First expand $(1+x)^{-n} = \bigg(\frac{1}{1-(-x)}\bigg)^n = (1 - x + x^2 - x^3 + \dots)^n$. Now, the coefficient on $x^k$ in that product is simply the number of ways to write $k$ as a sum of $n$ nonnegative numbers. That set of sums is in bijection to the set of diagrams with $k$ stars with $n-1$ bars among them. (For example, suppose $k=9$ and $n=4$. Then, **|*|***|*** corresponds to the sum $9=2+1+3+3$; ****||***|**corresponds to the sum $9 = 4+0+3+2$; ****|***||** corresponds to $9=4+3+0+2$; etc.) In each of these stars-and-bars diagrams we have $n+k-1$ objects, and we choose which ones are the $k$ stars in $\binom{n+k-1}{k}$ many ways. The $(-1)^k$ term comes from the alternating signs, and that proves the sum.

share|cite|improve this answer
This is exactly the sort of explanation I was looking for. I wanted to find one in terms of stars and bars: you rock. Also, the phrase "bijection to the set of diagrams with stars and bars" is epic. I want to say that all day long. I was wondering, though (and I guess this is another question entirely): why is (1 + x)^-n the number of ways to write k as a sum of n integers? What happens to the negative terms? – Ziggy Oct 28 '12 at 22:52
How do you know that the coefficient is indeed the number of n-tuples of non-negative integers whose sum is k? Can you give more details in this, please? – Ana Dec 2 '14 at 2:41

For a=1, the negative binomial series simplifies to (x+1)^(-n)=1-nx+1/2n(n+1)x^2-1/6n(n+1)(n+2)x^3+....

share|cite|improve this answer
Please rewrite your answer using TeX (…). – Ludolila Aug 22 '14 at 15:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.